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2 years ago

Which statement describes the translation from the graph of y = |x| to the graph of y = | x - 1 | -5

Mathematics

1 answer:

Doss [256]2 years ago

3 0

Answer:

Step-by-step explanation:

Shift 1 unit right and 5 units down

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#1. Which of the following suffixes means “bearing or producing”? A. -osis B. -oferous C. -fic D. -oid....#2. Which of the follo

AlexFokin [52]

#1-fic
#2 -oid

I took latin suffixes

5 0

2 years ago

Simplify the radical expression. ⁴√81x²⁰y²⁴<br><br>I think it's either A or B

tangare [24]

Answer:

3x^5y^6 (a)

Step-by-step explanation:

La raiz de un productor es igual al producto de las raices de cada factor

^4√81 ^4√x^20 ^4√y^24

escriba el numero en forma exponencial an base 3

^4√3^4 ^4√x^20 ^4√y^24

simplifica el indice de la raiz y el exponente usando 4

^4√3^4 x^5 ^4√y^24

simplifica el indice de la raiz y el exponente usando 4

^4√3^4 x^5 y^6

simplifica el indice de la raiz y el exponente usando 4

3x^5y^6

7 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

Aldo caught a 5 1/2 pound fish. How much did it weigh in ounces? answer in whole number or mixed number in simplest form.

Anarel [89]

1 pound is 16 ounces, so $5\frac{1}{2} lb * 16=80\frac{16}{2} oz=88oz$ .

3 0

3 years ago

Which matrix represents the system of equations shown below?

Leona [35]

Answer:

121-22

Step-by-step explanation:

7 0

3 years ago